Question

A STONE is thrown vertically upward with the velocity of25m/s .How long does it take to reach the maximum height ?Also calculate the height.

Answer 1

Answer:

time to Max height = 2.5 s

maxi height = 31.25 m

Explanation:

initial velocity = u = 25 m/s

final velocity = v = 0........ (at max height)

acceleration = a = -10 m/sq-sec

a/c eqn of motion,

v = u + at

=> t = (v - u)/a = -25/(-10) = 2.5 sec

again,

v^2 = u^2 + 2as

=> s = (v^2 - u^2)/2a = - 25 × 25 / (-2 × 10) = 31.25 m

max height = 31.25 m